 So what can you do with a definite integral? Well, you can find the area of a region, and that's pretty much our standard basic application of it, but there's a lot more you can do with it. So let's remind ourselves what the fundamental theorem of calculus is. If I have a definite integral, I can express it in the following way. And the things to note about this formula are definite integral is equal to f of b minus f of a. Well, that's just the change in a function value over some interval. So if nothing else, we can read the definite integral as telling us the change in some amount. What's the amount? Well, the other part of the fundamental theorem says that capital F prime of x is f of x. In other words, this function is, if we differentiate it, going to give us f of x. So f of x is the rate of change of our capital F function. So our integrand is going to be the rate of change of some quantity. And so that tells us that what our definite integral gives us is it gives us the change of a quantity where we know the rate of change of the quantity. So how does that translate into practical terms? Well, so let's take an example. The velocity of an object is modeled by 100 minus 32 feet per second, t seconds after it begins moving. How far does the object travel from t equals 0 to t equals 3? And so since this is a calculus class, we have questions about derivatives and tangent lines and definite integrals. And not one word of those appears anywhere in this question. So even though we've learned about these things, the question doesn't ask us to find a derivative. It does not ask us to find a definite integral. How do we solve this? Well, a good fallback is we might consider how to figure out what quantity is actually relevant. And a useful way of doing this goes back to this idea of dimensional analysis. And so let's supply some dimensional analysis to see if we can identify a quantity that might be relevant. Well, this is calculus. So there's two obvious quantities here, t, v of t. And since this is calculus, the derivative of v, and since we just learned about it, let's throw this in the integral of v of t, d, t, where I'm going to leave the limits of integration unstated right now. And the virtue of dimensional analysis is we can look at the dimensions of these quantities. And if the dimensions don't answer our question, how far? If the dimensions aren't relevant to that answer, we know the quantity itself is not what we're looking for. So let's go through those one at a time. So t is time measured in seconds. And so that is something that's measured in seconds. Definitely not going to be something that answers the question, how far? v of t, well, there's our function, it's measured in feet per second, feet over seconds. And that sounds more like a velocity. In fact, we're told that it's velocity. But that does not answer the question, how far? So again, this is not going to be relevant. v prime of t, that's the derivative of v of t. So remember how that's going to be measured, that's going to be units of v of t, feet per second, over units of t, seconds. And that's going to work out to be feet per second squared. Now, you might recognize that this is a unit of acceleration, and that's great because it tells you that if the problem asks you about acceleration, this is what you want. On the other hand, the important thing is that you should recognize this as not an answer to how far. Whatever you measure how far in, you know it's not feet per second squared. Well, I hope this definite integral of v of t, dt actually answers the question, how far? Because it's the only thing we have left. But let's not just rely on hope, let's look at the units that we're measuring the quantity in. So this integral of v of t, dt is going to be measured in feet, and so that tells me that this could possibly be an answer to the question, how far does an object travel? So let's take a look at that definite integral. So here's my v of t, dt, which tells me that I have to express the limits of integration in terms of t, and if only I had some way of knowing what those t values were. How about that? We know t equals 0 to t equals 3, so I'll substitute those in as my limits. Now, if this were an area problem, we'd be concerned that we don't have the right top and bottom curves, because we need to make sure that when we find an area, it turns out to be a non-negative number. However, the definite integral is much more than that. It measures an amount of change, and it's possible for that change to be positive or negative or zero. So we can just compute the actual numerical value of our definite integral. So I find an antiderivative. I evaluate it at the top limit and at the bottom limit, and I find the difference 156, and I should not forget to include the units there. We've already determined that the units are going to be in feet, and what that suggests is that if I want to find the distance, the only distance that I could possibly find, given this information in the problem, the only distance that I could find is 156 feet, and this is going to be the change in how far we've gone from t equals 0 to t equals 3, which makes a perfectly good answer to this question.